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poj 2155 (二維樹狀陣列 區間修改 求某點值)

Matrix
Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 33682   Accepted: 12194

Description

Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N). 

We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions. 

1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2). 
2. Q x y (1 <= x, y <= n) querys A[x, y]. 

Input

The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case. 

The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above. 

Output

For each querying output one line, which has an integer representing A[x, y]. 

There is a blank line between every two continuous test cases. 

Sample Input

1
2 10
C 2 1 2 2
Q 2 2
C 2 1 2 1
Q 1 1
C 1 1 2 1
C 1 2 1 2
C 1 1 2 2
Q 1 1
C 1 1 2 1
Q 2 1

Sample Output

1
0
0
1

#include<iostream>
#include
<cstdio> #include<cstring> #include<string> using namespace std; const int N=1e3+10; int s[N][N]; int n; int lowbit(int x) { return x&(-x); } void updata(int x,int y,int z) { for(int i=x;i<=n;i+=lowbit(i)){ for(int j=y;j<=n;j+=lowbit(j)){ s[i][j]+=z; } } } int sum(int x,int y) { int res=0; for(int i=x;i>0;i-=lowbit(i)){ for(int j=y;j>0;j-=lowbit(j)){ res+=s[i][j]; } } return res; } int main() { int T; scanf("%d",&T); while(T--){ memset(s,0,sizeof(s)); int m; scanf("%d %d",&n,&m); while(m--){ char t[3]; scanf("%s",t); if(t[0]=='C'){ int x1,y1,x2,y2; scanf("%d %d %d %d",&x1,&y1,&x2,&y2); updata(x1,y1,1); updata(x2+1,y2+1,1); updata(x2+1,y1,-1); updata(x1,y2+1,-1); } else{ int x,y; scanf("%d %d",&x,&y); printf("%d\n",sum(x,y)%2); } } if(T) printf("\n"); } return 0; }